Examples of omega-6 in the following topics:
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- $\displaystyle x_0 = -\frac{e E_0}{m} \frac{1}{\omega^2-\omega_0^2 - i \tau \omega_0^3} $
- $\displaystyle \tan \delta = \frac{\omega^3\tau}{\omega^2-\omega_0^2}~ \text{ and }~|x_0| = \frac{e E_0}{m} \left [ \left ( \omega^2-\omega_0^2 \right )^2 + \omega_0^6 \tau^2 \right ]^{-1/2}.$
- $\displaystyle P = \frac{q^2 |x_0|^2 \omega^4}{3 c^3} = \frac{q^4 E_0^2}{3m^2 c^3} \frac{\omega^4}{\left (\omega^2-\omega_0^2\right )^2+\left(\omega_0^3 \tau\right)^2}$
- $\omega \ll\ \omega_0: \sigma(\omega) \rightarrow \sigma_T \left (\frac{\omega}{\omega_0} \right )^4 $
- $\omega^2 - \omega_0^2 = (\omega - \omega_0) (\omega + \omega_0) \approx 2 \omega_0 (\omega-\omega_0)$
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- $\displaystyle \frac{dN}{dA dt} = 4\pi \frac{J_\nu}{\hbar \omega} d\nu = 4\pi \frac{J_\nu}{2\pi \hbar\omega} d\omega$
- $\displaystyle d\sigma = \frac{8\pi^2}{3 \hbar^2 c} \frac{\hbar\omega}{2 d\omega} | {\bf d}_{if} |^2 \left [ \frac{dn}{dp d\Omega} dp d\Omega \right ] $
- $\displaystyle \frac{d\sigma}{d\Omega} = \frac{p V m \omega}{6\pi c \hbar^3} |{\bf d}_{if} |^2 .$
- Let's calculate the cross-section for a photon with $\bar{h}\omega \gg 13.6 Z^2$~eV to ionize a hydrogen-like ion from the ground state.
- $\displaystyle |{\bf d}_{if} |^2 = \frac{256\pi}{V} \left ( \frac{Z}{a_0} \right )^5 \left ( \frac{Z^2}{a_0^2} + q^2 \right )^{-6} e^2 q^2 \approx \frac{256\pi e^2}{V} \left ( \frac{Z}{a_0} \right )^5 q^{-10}.$
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- $(- \omega ^2 + 2 \omega _0 ^2)A = \omega _0 ^2 B$
- $(- \omega ^2 + 2 \omega _0 ^2)B = \omega _0 ^2 A.$
- $\displaystyle{ (2 \omega _0 ^2 - \omega ^2)B = \frac{\omega _0 ^4} { 2 \omega _0 ^2 - \omega ^2 }B. }$
- $\displaystyle{ 2 \omega _0 ^2 - \omega ^2 = \omega _0 ^2 \Rightarrow \omega ^2 = \omega _0 ^2 . }$
- $\displaystyle{ 2 \omega _0 ^2 - \omega ^2 = - \omega _0 ^2 \Rightarrow \omega ^2 = 3 \omega _0 ^2 . }$
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- Even before calculating the form of $F(\omega/\omega_c)$, we can determine some interesting properties of the radiation spectrum.
- $\displaystyle P_\mbox{tot} (\omega) = C \int_{\gamma_1}^{\gamma_2} P(\omega) \gamma^{-p} d\gamma \propto \int_{\gamma_1}^{\gamma_2} F\left(\frac{\omega}{\omega_c}\right) \gamma^{-p}d\gamma.$
- Let's change variables to $x\equiv \omega/\omega_c$.
- Remember that $\omega_c = A \gamma^2$ so $\gamma^2 \propto \omega/x$, we get
- This power-law spectrum is valid essentially between $\omega_c(\gamma_1)$ and $\omega_c(\gamma_2)$.
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- $\displaystyle -\omega^2 \hat{\bf d}(\omega) = -\frac{e}{2\pi} \int_{-\infty}^\infty \dot{\bf v} e^{i\omega t} dt.$
- $\displaystyle \hat{\bf d}(\omega) \sim \left \{ \begin{array}{cl} \frac{e}{2\pi\omega^2} {\bf \Delta v}, &\omega \tau \ll\ 1 \\ 0, & \omega\tau \gg 1 \end{array} \right .$
- $\displaystyle {W(b)}{\omega} = \left \{ \begin{array}{cl} \frac{8 Z^2 e^6}{3\pi c^3 m^2 v^2 b^2}, & b \ll v/\omega \\ 0, & b \gg v/\omega \end{array} \right .$
- $\displaystyle \frac{d W}{d\omega dV dt} = n_e n_i 2\pi v \int_{b_\mbox{min}}^{b_\mbox{max}} \frac{8 Z^2 e^6}{3\pi c^3 m^2 v^2 b^2} b d b \\ = \frac{16 e^6}{3 c^3 m^2 v} n_e n_i Z^2 \int_{b_\mbox{min}}^{b_\mbox{max}} \frac{db}{b}\\ = \frac{16 e^6}{3 c^3 m^2 v} n_e n_i Z^2 \ln \left (\frac{b_\mbox{max}}{b_\mbox{min}} \right )$
- $\displaystyle \frac{d W}{d\omega dV dt} = \frac{16 \pi e^6}{3 \sqrt{3} c^3 m^2 v} n_e n_i Z^2 g_{ff}(v,\omega)$
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- $\displaystyle \frac{d W}{d\omega d\Omega} = \frac{q^2 \omega^2}{4\pi^2 c} \left | \int {\bf n} \times ({\bf n} \times \beta) \exp \left [ i \omega \left ( t'-{\bf n} \cdot {\bf r}_0 (t') / c \right ) \right ] dt' \right |^2$
- $t' - \frac{{\bf n} \cdot {\bf r}(t')}{c} = t' - \frac{a}{c} \cos \theta \sin \left ( \frac{v t'}{a} \right ) \\ \approx t' - \frac{a}{c} \left ( 1 - \frac{\theta^2}{2} \right ) \left [ \frac{v t'}{a} - \frac{1}{6} \left ( \frac{v t'}{a} \right )^3 \right ] \\ \approx t' - \beta t' + \frac{\theta^2}{2} \beta t' + \frac{c^2 \beta^3 t'^3}{6 a^2} \\ \approx \left ( 2 \gamma^2 \right )^{-1} \left [ \left ( 1 + \gamma^2 \theta^2 \right ) t' + \frac{c^2 \gamma^2 t'^3}{3a^2} \right ]$
- To get the final equation we took $x=\omega/\omega_c$ and $\beta \rightarrow 1$.
- $\displaystyle {W}{\omega d\Omega} \equiv {W_\|}{\omega d\Omega} + {W_\perp}{\omega d\Omega} \\ {W_\perp}{\omega d\Omega} = \frac{q^2 \omega^2}{4\pi^2 c} \left | \int \frac{ct'}{a} \exp \left [ \frac{i\omega}{2\gamma^2} \left ( \theta^2_\gamma t' + \frac{c^2 \gamma^2 t'^3}{3a^2} \right ) \right ] dt' \right |^2\\ {W_\|}{\omega d\Omega} = \frac{q^2 \omega^2 \theta^2}{4\pi^2 c} \left | \int \exp \left [ \frac{i\omega}{2\gamma^2} \left ( \theta^2_\gamma t' + \frac{c^2 \gamma^2 t'^3}{3a^2} \right ) \right ] dt' \right |^2$
- $\displaystyle \frac{W_\perp}{\omega d\Omega} = \frac{q^2 \omega^2}{4\pi^2 c} \left ( \frac{a \theta^2_\gamma}{\gamma^2 c} \right )^2 K^2_\frac{2}{3} (\eta)\\ \frac{W_\|}{\omega d\Omega} = \frac{q^2 \omega^2 \theta^2}{4\pi^2 c} \left ( \frac{a \theta_\gamma}{\gamma c} \right )^2 K^2_\frac{1}{3} (\eta)$
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- $\displaystyle {\hat E}_x(\omega) = i \frac{q}{\pi \gamma b v} \left [ \frac{\omega b}{\gamma v} K_0 \left ( \frac{\omega b}{\gamma v} \right )\right ]$
- $\displaystyle \\ {\hat E}_y(\omega) = \frac{q}{\pi b v} \left [ \frac{\omega b}{\gamma v} K_1 \left ( \frac{\omega b}{\gamma v} \right ) \right ].$
- $\\ {\hat E}_y(\omega) = \frac{q}{\pi b v} \left [ \frac{\omega b}{\gamma v} K_1 \left ( \frac{\omega b}{\gamma v} \right ) \right ].$
- $\displaystyle \frac{dW}{dAd\omega} = c |{\hat E}(\omega)|^2 = \left \{ \begin{array}{cl} \frac{ q^2 c}{\pi^2 v^2 b^2}, & b \ll \gamma v/\omega \\ 0, & b \gg \gamma v/\omega \end{array} \right .$
- $\displaystyle \frac{dW}{d\omega} = \sigma_T \frac{\sigma(\omega)}{\sigma_T} \frac{dW}{dAd\omega} = \left \{ \begin{array}{cl} \frac{ 8 \pi Z^2 e^6}{3 v^2 m^2 c^3 b^2} \frac{\sigma(\omega)}{\sigma_T}, & b \ll \gamma v/\omega \\ 0, & b \gg \gamma v/\omega \end{array} \right .$
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- Defining $\omega_0 ^2 = k/m$ and $\Omega ^2 = k'/m$ , we have $\omega _ + ^2 = \omega _0 ^2$ and $\omega _ - ^2 = \omega _0 ^2 + 2 \Omega ^2 $ .
- Answer: $mg = kx$ , so $x = mg/k = .1 {\rm kg}~ 9.8 {\rm \frac{m}{s^2}}/ 15.8 {\rm \frac{N}{m}} = .06 {\rm m} = 6 {\rm cm}$ .
- 1.6 Now suppose that the mass is sitting in a viscous fluid.
- $(-\omega ^2 + 2 \omega_1 ^2) (-\omega ^2 + 2 \omega_2 ^2) = \omega _1 ^2 \omega _2 ^2.$
- $\omega _{\pm} ^2 = \omega _1 ^2 + \omega _2 ^2 \pm \omega_1 ^2$
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- $\displaystyle{ ( -\omega^2 + i \gamma \omega + \omega _ 0 ^2) \hat{x} e^{i \omega t} = \frac{ \hat{F}}{m} e^{i \omega t}.
- $\displaystyle{ \hat{x} = \frac{ \hat{F} /m} { -\omega^2 + i \gamma \omega + \omega _ 0 ^2}. }$
- $\displaystyle{ -\omega^2 + i \gamma \omega + \omega _ 0 ^2 = \sqrt{ (\omega _ 0 ^2 -\omega^2 )^2 + \gamma^2 \omega ^2} e ^ {i \tan ^{-1} \frac{\gamma \omega}{ \omega _ 0 ^2 -\omega^2 }}. }$
- The function $\rho ^2 = 1/ ((\omega _ 0 ^2 -\omega^2 )^2 + \gamma^2 \omega ^2)$ has a characteristic shape seen in all resonance phenomena.
- It's peaked about the characteristic frequency $\omega_0$ and has a full width of $\gamma$ at half its maximum height as illustrated in Figure 1.6.
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- $\displaystyle \int_{-\infty}^{\infty} |E(t)|^2 dt = 2\pi \int_{-\infty}^{\infty} |{\hat E}(\omega)|^2 d \omega.$
- $\displaystyle \int_{-\infty}^{\infty} |E(t)|^2 dt = \int_{-\infty}^{\infty} d t \int_{-\infty}^{\infty} {\hat E}(\omega') e^{-i\omega' t} d \omega'.
- \int_{-\infty}^{\infty} {\hat E}^*(\omega) e^{i\omega t} d \omega \\ \displaystyle = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} d t d\omega' d \omega {\hat E}(\omega') {\hat E}^*(\omega) e^{-i\omega' t} e^{i\omega t} $
- The integral over time is simply Fourier transform of $2\pi e^{-i\omega' t}$ which we know,
- $\displaystyle \int_{-\infty}^{\infty} |E(t)|^2 dt = 2\pi \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} d\omega' d \omega {\hat E}(\omega') {\hat E}^*(\omega) \delta (\omega -\omega') \\ \displaystyle = 2 \pi \int_{-\infty}^{\infty} d \omega {\hat E}(\omega) {\hat E}^*(\omega) =2 \pi \int_{-\infty}^{\infty} |{\hat E}(\omega)|^2 d \omega$